Unformatted text preview: τ 1 = (2.00m) F 2 – (1.00m) 196N – (1.70m) 490.N = I α = 0 , so (2.00m) F 2 = 196N m + 833N m = 1029N m , so F 2 = (1029N m) / (2.00m) = F 2 = 515N . Moving the origin to the right support we have the net torque is given by τ 2 = (2.00m) F 1 + (1.00m) 196N + (0.30m) 490.N = I α = 0 , so (2.00m) F 1 = 196N m + 147N m = 343N m , so F 1 = (343N m) / (2.00m) = F 1 = 172N . Notice that F 1 + F 2 = W pl + W pe (just off by one at the last sig. fig.). F 1 F 2 W pl W pe 1.70m...
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This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.
 Spring '11
 lomant

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